The image shows a full-wave rectifier circuit with two diodes (D1, D2), a transformer with a turn ratio of 18:1, a capacitor $C$, and a load resistor $R_L$. The power supply voltage is $V_s = 220 \, V$. The problem is split into two parts:

Problem (a):

• Given:
  • $R_L = 10 \, \Omega$,
  • $C_1 = 1000 \, \mu F$,
  • $C_2 = 2000 \, \mu F$.

The task is to determine the output voltage $V_{LDC}$.

Problem (b):

• Given:
  • $C = 1000 \, \mu F$,
  • $R_L = 10 \, \Omega$,
  • $R_L = 100 \, \Omega$.

The task is to calculate the ripple factor $r$ in percentage.

Step-by-Step Solution Outline:

For part (a):

1. Peak Secondary Voltage Calculation: The transformer reduces the input voltage to the secondary side by a factor of 18: $V_{peak} = \frac{V_s}{18 \sqrt{2}} = \frac{220}{18 \sqrt{2}} \, V$ This gives the peak value of the rectified voltage before smoothing.

2. DC Output Voltage: The capacitor smooths the rectified signal. The approximate DC output voltage $V_{LDC}$ is: $V_{LDC} \approx V_{peak} - V_{diode-drop}$ where $V_{diode-drop} \approx 0.7 \, V$ for each diode.

For part (b):

1. Ripple Voltage: The ripple voltage $V_{ripple}$ for a full-wave rectifier is given by: $V_{ripple} = \frac{I_{load}}{f C}$ where:

   • $I_{load} = \frac{V_{LDC}}{R_L}$,
   • $f$ is the frequency of the ripple (which is twice the mains frequency for full-wave rectification, typically $2 \times 50 = 100 \, Hz$).

2. Ripple Factor: The ripple factor $r$ is calculated as: $r = \frac{V_{ripple}}{V_{LDC}} \times 100 \%$

Would you like the full numeric calculations for each part?

Related Questions:

1. What effect does increasing the capacitance have on the ripple factor?
2. How does the load resistance impact the output voltage?
3. Why is the frequency of ripple in a full-wave rectifier twice the mains frequency?
4. What would happen to $V_{LDC}$ if the diode drop were larger (e.g., with different diodes)?
5. How do you choose an appropriate capacitor for a rectifier circuit?

Tip:

Larger capacitance in the smoothing capacitor reduces the ripple voltage but increases charging current, which may affect diode stress.